@Dave: Dude...u didn't count the cost for tying the two ropes i.e. (3 and 4)
together with (5 and 6).

On Mon, Mar 28, 2011 at 9:19 PM, Dave <[email protected]> wrote:

> @Bittu. The "ungreedy" algorithm works. Repeatedly tie the two
> shortest ropes.
>
> E.g., suppose the ropes are 3, 4, 5, and 6 units long. Then tie 3 and
> 4, giving 7. Now 5 and 6 are the two shortest, so tie them, giving 11.
> Finally, 7 + 11 = 18.
>
> Dave
>
> On Mar 28, 1:41 am, bittu <[email protected]> wrote:
> > you are given n ropes,maybe of different length. the cost of tying two
> > ropes is the sum of their lengths.Find a way to tie these ropes
> > together so that the cost is minimum.
> >
> > Thanks
> > Shashank
>
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